Simplify and expand the following expression: $ \dfrac{3}{n + 9}- \dfrac{3}{4n + 16}+ \dfrac{3}{n^2 + 13n + 36} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4n + 16} = \dfrac{3}{4(n + 4)}$ We can factor the quadratic in the third term: $ \dfrac{3}{n^2 + 13n + 36} = \dfrac{3}{(n + 9)(n + 4)}$ Now we have: $ \dfrac{3}{n + 9}- \dfrac{3}{4(n + 4)}+ \dfrac{3}{(n + 9)(n + 4)} $ The least common multiple of the denominators is: $ (n + 9)(n + 4)$ In order to get the first term over $(n + 9)(n + 4)$ , multiply by $\dfrac{4(n + 4)}{4(n + 4)}$ $ \dfrac{3}{n + 9} \times \dfrac{4(n + 4)}{4(n + 4)} = \dfrac{12(n + 4)}{(n + 9)(n + 4)} $ In order to get the second term over $(n + 9)(n + 4)$ , multiply by $\dfrac{n + 9}{n + 9}$ $ \dfrac{3}{4(n + 4)} \times \dfrac{n + 9}{n + 9} = \dfrac{3(n + 9)}{(n + 9)(n + 4)} $ In order to get the third term over $(n + 9)(n + 4)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{3}{(n + 9)(n + 4)} \times \dfrac{4}{4} = \dfrac{12}{(n + 9)(n + 4)} $ Now we have: $ \dfrac{12(n + 4)}{(n + 9)(n + 4)} - \dfrac{3(n + 9)}{(n + 9)(n + 4)} + \dfrac{12}{(n + 9)(n + 4)} $ $ = \dfrac{ 12(n + 4) - 3(n + 9) + 12} {(n + 9)(n + 4)} $ Expand: $ = \dfrac{12n + 48 - 3n - 27 + 12}{4n^2 + 52n + 144} $ $ = \dfrac{9n + 33}{4n^2 + 52n + 144}$